
282 Chapter 7 Evolutionary inverse problems
with given α
0
= 0. and q = 0.75. To calculate the discrepancy, we additionally solved
the direct problem in which initial data were taken equal to the solution of the inverse
problem.
For simplicity, in the program presented below we have restricted ourselves to the
simplest case of time-independent coefficients, with k(x) = 1 in (7.77), and the
weighting parameter in the regularized additive scheme (7.155), (7.156) is σ = 1.
Program PROBLEM11
C
C PROBLEM11 INVERTED-TIME PROBLEM
C TWO-DIMENSIONAL PROBLEM
C ADDITIVE REGULARIZED SCHEME
C
IMPLICIT REAL
*
8 ( A-H, O-Z )
PARAMETER ( DELTA = 0.01, N1 = 101, N2 = 101, M = 101 )
DIMENSION U0(N1,N2), UT(N1,N2), UTD(N1,N2), U(N1,N2)
+ ,X1(N1), X2(N2), Y(N1,N2), Y1(N1,N2), Y2(N1,N2), YY(N1)
+ ,A(N1), B(N1), C(N1), D(N1), E(N1), F(N1) ! N1 >= N2
C
C PARAMETERS:
C
C X1L, X2L - COORDINATES OF THE LEFT CORNER;
C X1R, X2R - COORDINATES OF THE RIGHT CORNER;
C N1, N2 - NUMBER OF NODES IN THE SPATIAL GRID;
C H1, H2 - STEP OVER SPACE;
C TAU - TIME STEP;
C DELTA - INPUT-DATA INACCURACY;
C Q - MULTIPLIER IN THE REGULARIZATION PARAMETER;
C U0(N1,N2) - RECONSTRUCTED INITIAL CONDITION;
C UT(N1,N2) - END-TIME SOLUTION OF THE DIRECT PROBLEM;
C UTD(N1,N2)- DISTURBED SOLUTION OF THE DIRECT PROBLEM;
C U(N1,N2) - APPROXIMATE SOLUTION OF THE INVERSE PROBLEM;
C
X1L = 0.D0
X1R = 1.D0
X2L = 0.D0
X2R = 1.D0
TMAX = 0.025D0
PI = 3.1415926D0
C
OPEN (01, FILE = ’RESULT.DAT’) !FILE TO STORE THE CALCULATED DATA
C
C GRID
C
H1 = (X1R-X1L) / (N1-1)
H2 = (X2R-X2L) / (N2-1)
TAU = TMAX / (M-1)
DOI=1,N1
X1(I) = X1L + (I-1)
*
H1
END DO
DOJ=1,N2
X2(J) = X2L + (J-1)
*
H2
END DO
C
C DIRECT PROBLEM